Carnot Efficiency Formula
The efficiency ($\eta$) of a Carnot engine depends on the hot ($T_H$) and cold ($T_C$) reservoir temperatures:
$ \eta = 1 - \frac{T_C}{T_H} $
Engine A Efficiency ($\eta_A$)
Engine A operates between $T_{HA} = 600$ K and rejects heat to a reservoir at temperature $T$. Its efficiency ($\eta_A$) is:
$ \eta_A = 1 - \frac{T}{600} $
Engine B Efficiency ($\eta_B$)
Engine B receives the heat rejected by Engine A (at temperature $T$) and rejects heat to a reservoir at $T_{CB} = 100$ K. Its efficiency ($\eta_B$) is:
$ \eta_B = 1 - \frac{100}{T} $
Efficiency Ratio Calculation
The question asks for the ratio $\frac{\eta_B}{\eta_A}$. Substituting the efficiency expressions:
$ \frac{\eta_B}{\eta_A} = \frac{1 - \frac{100}{T}}{1 - \frac{T}{600}} $
We are given options for this ratio. Using the correct answer choice $\frac{7}{12}$, we set the ratio equal to this value:
$ \frac{1 - \frac{100}{T}}{1 - \frac{T}{600}} = \frac{7}{12} $
Intermediate Temperature $T$ Determination
To find the value of the intermediate temperature $T$, we solve the equation:
$ 12 \left( 1 - \frac{100}{T} \right) = 7 \left( 1 - \frac{T}{600} \right) $
$ 12 - \frac{1200}{T} = 7 - \frac{7T}{600} $
Rearrange the terms to solve for $T$:
$ 5 = \frac{1200}{T} - \frac{7T}{600} $
Multiply the entire equation by $600T$ to eliminate denominators:
$ 5 \times 600T = 1200 \times 600 - 7T^2 $
$ 3000T = 720000 - 7T^2 $
This yields a quadratic equation:
$ 7T^2 + 3000T - 720000 = 0 $
Using the quadratic formula ($T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$), we find the positive temperature solution:
$ T = \frac{-3000 + \sqrt{3000^2 - 4(7)(-720000)}}{2(7)} = \frac{-3000 + \sqrt{29160000}}{14} = \frac{-3000 + 5400}{14} = \frac{2400}{14} $
$ T = \frac{1200}{7} \text{ K} $
Verification of Ratio
With $T = \frac{1200}{7}$ K, we calculate the individual efficiencies:
$ \eta_A = 1 - \frac{1200/7}{600} = 1 - \frac{2}{7} = \frac{5}{7} $
$ \eta_B = 1 - \frac{100}{1200/7} = 1 - \frac{7}{12} = \frac{5}{12} $
Now, calculate the ratio $\frac{\eta_B}{\eta_A}$:
$ \frac{\eta_B}{\eta_A} = \frac{5/12}{5/7} = \frac{5}{12} \times \frac{7}{5} = \frac{7}{12} $
This result matches the correct answer.
Similar Questions
10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_1$ to $P_2$ is $\alpha \text{ Joule}$ ($P_1 = 21.7 \text{ Pa}$ and $P_2 = 30 \text{ Pa}, C_v = 21 \text{ J/K.mol}, R = 8.3 \text{ J/mol.K}$). The value of $\alpha$ is _______.
- When $300 \text{ J}$ of heat given to an ideal gas with $C_p = \frac{7}{2} R$ its temperature raises from $20^\circ\text{C}$ to $50^\circ\text{C}$ keeping its volume constant. The mass of the gas is (approximately) _______ g. ($R = 8.314 \text{ J/mol.K}$)
A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle $ABC$ is _________ $\text{J}$.
- The heat extracted out of $x$ gram of water initially at $50 \ ^{\circ}\text{C}$ to cool it down to $0 \ ^{\circ}\text{C}$ is sufficient to evaporate $(1000 - x)$ gram of water also initially at $50 \ ^{\circ}\text{C}$. The value of $x$ (closest integer) is ___________.
(Take latent heat of water $2256 \text{ kJ/kg. K}$, specific heat capacity of water $4200 \text{ J/kg. K}$) - A vessel contains $0.15\text{ m}^3$ of a gas at pressure $8\text{ bar}$ and temperature $140^{\circ}\text{C}$ with $c_p = 3R$ and $c_v = 2R$. It is expanded adiabatically till pressure falls to $1\text{ bar}$. The work done during this process is ________ $\text{kJ}$. (R is gas constant)
- The r.m.s. speed of oxygen molecules at $47^\circ\text{C}$ is equal to that of the hydrogen molecules kept at __________ $^\circ\text{C}$. (Mass of oxygen molecule/mass of hydrogen molecule $= 32/2$)
- A Carnot engine is operating between the two reservoirs of temperature $227 \ °C$ and $127 \ °C$ respectively. If the engine performs $1.04 \ \times \ 10^5 \ J$ work per cycle. Then, the heat absorbed from hot reservoir per cycle will be :
- A body takes 10 minutes to cool from $60^\circ C$ to $50^\circ C$. The temperature of surroundings is constant at $25^\circ C$. Then, the temperature of the body after next 10 minutes will be approximately :
- The value closest to the thermal velocity of a Helium atom at room temperature (300 K) in $ms^{-1}$ is : [$k_B = 1.4 \times 10^{-23}$ J/K; $m_{He} = 7 \times 10^{-27}$ kg]
- In 5 minutes, a body cools from $75^{\circ}C$ to $65^{\circ}C$ at room temperature of $25^{\circ}C$. The temperature of body at the end of next 5 minutes is __________ $^{\circ}C$.
Important Questions from Heat and Thermodynamics
10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_1$ to $P_2$ is $\alpha \text{ Joule}$ ($P_1 = 21.7 \text{ Pa}$ and $P_2 = 30 \text{ Pa}, C_v = 21 \text{ J/K.mol}, R = 8.3 \text{ J/mol.K}$). The value of $\alpha$ is _______.
- When $300 \text{ J}$ of heat given to an ideal gas with $C_p = \frac{7}{2} R$ its temperature raises from $20^\circ\text{C}$ to $50^\circ\text{C}$ keeping its volume constant. The mass of the gas is (approximately) _______ g. ($R = 8.314 \text{ J/mol.K}$)
A thermodynamic system is taken through the cyclic process ABC as shown in the figure. The total work done by the system during the cycle $ABC$ is _________ $\text{J}$.
- The heat extracted out of $x$ gram of water initially at $50 \ ^{\circ}\text{C}$ to cool it down to $0 \ ^{\circ}\text{C}$ is sufficient to evaporate $(1000 - x)$ gram of water also initially at $50 \ ^{\circ}\text{C}$. The value of $x$ (closest integer) is ___________.
(Take latent heat of water $2256 \text{ kJ/kg. K}$, specific heat capacity of water $4200 \text{ J/kg. K}$) - Two identical metal bars are heated in two different temperatures and allowed to cool in the same surroundings. Which one of the following figures correctly shows their cooling curves?
